In the figure shown, A B is horizontal diameter of the vertical circle. Time taken by a particle to slide along chord A C is t1 and that corresponding to chord B C is t2. If t1 × t2=(x R/y), then find (x+y)

Question

In the figure shown, A B is horizontal diameter of the vertical circle. Time taken by a particle to slide along chord A C is t1 and that corresponding to chord B C is t2. If t1 × t2=(x R/y), then find (x+y) (A)  (B)  (C)  (D)

Tardigrade Admin · Accepted Answer

A C=2 R cos θ ; B C=2 R sin θ 2 R cos θ=(1/2)(g sin θ) t12 ...(i) 2 R sin θ=(1/2)(g cos θ) t22....(ii) <img class="img-fluid question-image" alt="image" src="https://cdn.tardigrade.in/img/question/physics/afd065fbb7ca4770b98b28045eef1e41-.png" /> From (i) and (ii), we get t2 / t1= tan θ and t1 × t2=4 R / 9

Q. In the figure shown, $A B$ is horizontal diameter of the vertical circle. Time taken by a particle to slide along chord $A C$ is $t_{1}$ and that corresponding to chord $BC$ is $t_{2}$ . If $t_{1} \times t_{2} = \frac{x R}{y}$ , then find $(x + y)$

$A C = 2 R cos θ; BC = 2 R sin θ$
$2 R cos θ = \frac{1}{2} (g sin θ) t_{1}^{2}$ ...(i)
$2 R sin θ = \frac{1}{2} (g cos θ) t_{2}^{2}$ ....(ii)

From (i) and (ii), we get
$t_{2} / t_{1} = tan θ$
and $t_{1} \times t_{2} = 4 R /9$

Q. In the figure shown, AB is horizontal diameter of the vertical circle. Time taken by a particle to slide along chord AC is t1​ and that corresponding to chord BC is t2​. If t1​×t2​=yxR​, then find (x+y)

AC=2Rcosθ;BC=2Rsinθ 2Rcosθ=21​(gsinθ)t12​ ...(i) 2Rsinθ=21​(gcosθ)t22​....(ii) From (i) and (ii), we get t2​/t1​=tanθ and t1​×t2​=4R/9

Q. In the figure shown, $A B$ is horizontal diameter of the vertical circle. Time taken by a particle to slide along chord $A C$ is $t_{1}$ and that corresponding to chord $BC$ is $t_{2}$ . If $t_{1} \times t_{2} = \frac{x R}{y}$ , then find $(x + y)$

$A C = 2 R cos θ; BC = 2 R sin θ$
$2 R cos θ = \frac{1}{2} (g sin θ) t_{1}^{2}$ ...(i)
$2 R sin θ = \frac{1}{2} (g cos θ) t_{2}^{2}$ ....(ii)

From (i) and (ii), we get
$t_{2} / t_{1} = tan θ$
and $t_{1} \times t_{2} = 4 R /9$